Q.The value of $\displaystyle\int_{0}^{a}\left(\sqrt{a^2-x^2}\right)^3\,dx$ is :

A reduction formula expresses a definite integral carrying an index (power) $n$, written $I_n$, in terms of the same type of integral with a smaller index ($I_{n-1}$ or $I_{n-2}$). Applying the formula repeatedly ("reducing" the index each time) eventually collapses the whole integral to an elementary base case ($I_0$ or $I_1$).

Method (3 steps). (1) Identify the index $n$. (2) Name the integral $I_n$. (3) Apply integration by parts once to relate $I_n$ to $I_{n-2}$ (or $I_{n-1}$); this relation is the reduction formula.

Reduction Formula I. If $I_n=\displaystyle\int_0^{\pi/2}\sin^nx\,dx$, then $I_n=\dfrac{n-1}{n}I_{n-2}$, $n\ge2$.

Reduction Formula II. If $I_n=\displaystyle\int_0^{\pi/2}\cos^nx\,dx$, then $I_n=\dfrac{n-1}{n}I_{n-2}$, $n\ge2$ (same recursion as sine, by symmetry).

Reduction Formula III. If $I_{m,n}=\displaystyle\int_0^{\pi/2}\sin^mx\cos^nx\,dx$, then $I_{m,n}=\dfrac{n-1}{m+n}I_{m,n-2}$, $n\ge2$ (reduces the cosine power; by symmetry the sine power can equally be reduced, and it is always convenient to reduce whichever of $m,n$ is odd first).

Reduction Formula IV. If $I_{m,n}=\displaystyle\int_0^1 x^m(1-x)^n\,dx$, then $I_{m,n}=\dfrac{n}{m+n+1}I_{m,n-1}$, $n\ge1$.

Closed (Wallis-type) forms, obtained by iterating I-III down to the base case: …